algebra precalculus - Does $(p^{lambda_1}-1)cdots(p^{lambda_m}-1)$ divide $p^n-1$?

Let $p$ be a prime integer and $n\geq 3$ be a natural number such that $p^n-1$ has a primitive prime divisor. Also let $n=\lambda_1+\cdots+\lambda_m$ such that $n>\lambda_i\geq 1$ for all $i$. I want to show that $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$ does not divide $p^n-1$.

This is my suggested argument for proving the statement when $p>2$:
Both of $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$ and $p^n-1$ are polynomials in $p$ with the same degree. If $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)\mid p^n-1$, we then have $p^n-1=c(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$, where $c\in \mathbb{N}$. Since $p>2$, $p^{\lambda_i}-1>1$ for all $i$. Therefore, the coefficient of $p^{\lambda_1+...+\lambda_m}=p^n$ in the expansion of the product $\prod_{i=1}^{m}{(p^{\lambda_i}-1)}$ is 1. Also the coefficient of $p^n$ in $p^n-1$ is $1$, and so we must have $c=1$ . Therefore $p^n-1=(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$. Now if we consider $q$ to be primitive prime divisor of $p^n-1$, then $q$ does not divide $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$, which is a contradiction. So if $p>2$, it is impossible for $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$ to divide $p^n-1$.

But as it was discussed in comments, it seems that my argument is not true.
Question 1: Does my argument work? If not, is there any possible way to prove the statement for $p>2$?
The remaining case is when $p=2$. According to the example mentioned in comments $(2-1)(2^2-1)(2^3-1) \mid 2^6-1$.
Question 2: Is there any example of a pair $(n, p)\neq(6,2)$ and a partition $\lambda=(1\leq\lambda_1\leq...\leq\lambda_mI would be grateful for any help.